How do you factor the trinomial 6a^2-54a+1086a2−54a+108?

1 Answer
Shwetank Mauria
Apr 7, 2016

Factors o 6a^2-54a+1086a2−54a+108 are 6*(a-6)*(a-3)6⋅(a−6)⋅(a−3)

Explanation:

In 6a^2-54a+1086a2−54a+108, each of the coefficients is divisible by 66, so taking 66 common we get

6a^2-54a+108=6xx(a^2-9a+18)6a2−54a+108=6×(a2−9a+18)

Now to factorize a^2-9a+18a2−9a+18 let us split middle term into components whose product is 1818 and these are 66 and 33.

Hence, 6a^2-54a+1086a2−54a+108 can be written as

6(a^2-6a-3a+18)6(a2−6a−3a+18) or

6*{a(a-6)-3(a-6}6⋅{a(a−6)−3(a−6} or

6*(a-6)*(a-3)6⋅(a−6)⋅(a−3)

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